Question

How do you solve the following limit `(e^x-1)/x` as x approaches zero?

Answer 1

It is a remarkable limit, but, if you want to demonstrate it, you have to know the fundamental limit:

`lim_(xrarroo)(1+1/x)^x=e` (number of Neper), and also this limit:

`lim_(xrarr0)(1+x)^(1/x)=e` that it is easy to demonstrate in this way:

let `x=1/t`, so when `xrarr0` than `trarroo` and this limit becomes the first one.

So:

let `e^x-1=trArre^x=t+1rArrx=ln(t+1)`

and if `xrarr0rArrtrarr0`

`lim_(xrarr0)(e^x-1)/x=lim_(trarr0)t/ln(t+1)=lim_(trarr0)1/(ln(t+1)/t)=`

`=lim_(trarr0)1/(1/tln(t+1))=lim_(trarr0)1/(ln(t+1)^(1/t))=`

(for the second limit)`=1/(lne)=1/1=1`.

Answer 2

The limit is in the indeterminate form `(e^0-1)/0=0/0`, so l'Hopital's rule applies:

`lim_(xrarr0)(e^x-1)/x=lim_(xrarr0)(d/dx(e^x-1))/(d/dx(x))=lim_(xrarr0)e^x/1=e^0=1`